小知识，大挑战！本文正在参与“程序员必备小知识”创作活动。
描述
There are n people that are split into some unknown number of groups. Each person is labeled with a unique ID from 0 to n - 1.
You are given an integer array groupSizes, where groupSizes[i] is the size of the group that person i is in. For example, if groupSizes[1] = 3, then person 1 must be in a group of size 3.
Return a list of groups such that each person i is in a group of size groupSizes[i].
Each person should appear in exactly one group, and every person must be in a group. If there are multiple answers, return any of them. It is guaranteed that there will be at least one valid solution for the given input.
Example 1:
Input: groupSizes = [3,3,3,3,3,1,3]
Output: [[5],[0,1,2],[3,4,6]]
Explanation:
The first group is [5]. The size is 1, and groupSizes[5] = 1.
The second group is [0,1,2]. The size is 3, and groupSizes[0] = groupSizes[1] = groupSizes[2] = 3.
The third group is [3,4,6]. The size is 3, and groupSizes[3] = groupSizes[4] = groupSizes[6] = 3.
Other possible solutions are [[2,1,6],[5],[0,4,3]] and [[5],[0,6,2],[4,3,1]].
Example 2:
Input: groupSizes = [2,1,3,3,3,2]
Output: [[1],[0,5],[2,3,4]]
Note:
groupSizes.length == n
1 <= n <= 500
1 <= groupSizes[i] <= n
解析
根据题意，有 n 个人被分成不知数量的几个组，每个人的唯一 ID 标记就是索引 0 到 n-1 ，每个人都应该恰好出现在一个组中，并且每个人都必须在一个组中。
然后又给了一个整数列表 groupSizes ，其中 groupSizes[i] 表示的是索引为 i 的人所在组的大小。例如，如果 groupSizes[1] = 3，则索引为 1 的人必须在大小为 3 的组中。
要求返回一个满足题意的列表，如果有多个答案，则返回其中任何一个。
其实这个题就是有点绕，理清了题意就好做了，这里主要是借用字典来解题：
-
初始化结果 result 为空列表存放答案， d 为空字典用来存放不同的 size 对应的索引列表，如例子一中得到的中间答案就是：
d = {1: [5], 3: [0, 1, 2, 3, 4, 6]} -
然后遍历 d 中的 k,v 对，v 是列表，且列表的长度肯定是 k 的整数倍，所以用下面的语句截取长度为 k 的列表放入 result 中：
for i in range(0,len(v),k): result.append(v[i:i+k]) -
遍历最后得到的 result 就是答案。
解答
class Solution(object):
def groupThePeople(self, groupSizes):
"""
:type groupSizes: List[int]
:rtype: List[List[int]]
"""
result = []
d = {}
c = collections.Counter(groupSizes)
for k,v in c.items():
d[k] = []
for i, size in enumerate(groupSizes):
d[size].append(i)
for k,v in d.items():
for i in range(0,len(v),k):
result.append(v[i:i+k])
return result
运行结果
Runtime: 68 ms, faster than 46.94% of Python online submissions for Group the People Given the Group Size They Belong To.
Memory Usage: 13.4 MB, less than 99.32% of Python online submissions for Group the People Given the Group Size They Belong To.
原题链接：leetcode.com/problems/gr…
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